Hi
I have a Ball.
- The Ball moves forward at an initial speed S ( pixels per frame ).
- Every Frame the Balls current Speed is multiplied by 0.99
I need to determine, ahead of time through an expression, how far the Ball has traveled when Ball has reached TS ( Target Speed; always lower than initial speed ).
...
How do i need to think to solve something like this?
How can i solve this?
"How do I need to think to solve something like this?"
If you're familiar with Newton's equations of motion (fairly standard school maths), you have a constant acceleration (actually a deceleration, since you're multiplying your speed by a value less than 1, in your case 0.99) so you can use the SUVAT equations
#3 seems to fit the bill:
s=0.5(u+v)t
where:
s=distance travelled (pixels in your example)
u=starting velocity (in your example S)
v=ending velocity (in your example TS)
t=time (in your case number of frames)
If you need to calculate t beforehand, you can use #1:
v=u+at
so t=v/(u+a)
a=acceleration (in your example -0.99, I think)
Hope that gets you on the right track!
Without any background in physics this would be hard to figure out.
...Maybe i should take up physics on the side...
Ill post back as soon as i get it working.
Thanks AyreGuitar!
Is Speed and Velocity the same thing?
"Velocity is a vector quantity that refers to "the rate at which an object changes its position.""
"Speed is a scalar quantity that refers to "how fast an object is moving.""
Either its not working or im doing something wrong.
https://dl.dropboxusercontent.com/u/...Prediction.mfa
Would it be significantly easier to predict Travel Distance at 50% of Initial Speed?
You're right it's not working.
I think I made an error in thinking that using speed=0.99*speed is using constant acceleration, some quick calculations reveal that its not - sorry about that!
Also I don't think -0.99 would have been your acceleration (it's more like -0.04, you can calculate it by using v=u+at, so a=(v-u)/t, a=(4*0.99-4)/1=-0.04)
I also should have been using SUVAT equation 4 in my link, so
s=(v*v-u*u)/2*a
If u=4, v=3 and a=-0.04 this results in s=(9-16)/2*(-0.04)=-7/-0.08=87.5 which is closer but a little off as a keeps decreasing:
frame 1, a=((4*0.99)-4)/1=-0.04
frame 2, a=((3.96*0.99)-4)/2=-0.0398
etc
To use these equations accurately, you'd need constant acceleration, ie speed=speed+Value rather than speed=speed*Value
If you need to use something like speed=speed*Value, you might be better off using a fast loop and calculating the new distance and speed until speed<=target speed
Sorry for the confusion!
I've attached an updated version of your example in case it helps (there was definitely something wrong with the expression as even with the new acceleration it still gave very small values, but I've re-written it to use a 'proper' version of acceleration)
BTW About Velocity and Speed, most of the time you can use them in the same way, the main thing is that velocity has a direction (Vector) and speed does not (Scalar), so if direction is important use velocity instead of speed.
Hmm, i see.
So if i want to comunicate a Velocity to someone, it wont make sense unless i comunicated a speed and a direction/ angle.
Thanks AyreGuitar!
...
I actually managed to find an expresion which is very accurate:
V_i = VelocityInitial
V_t = VelocityTarget
D = Deceleration
( ( 0.0 + V_i ) / Ln( D ) ) * ( D ^ ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )
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